| Exam Board | SPS |
|---|---|
| Module | SPS FM Mechanics (SPS FM Mechanics) |
| Year | 2025 |
| Session | January |
| Marks | 14 |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with direction reversal |
| Difficulty | Standard +0.3 This is a standard A-level mechanics collision problem requiring conservation of momentum and the coefficient of restitution formula. Part (a) involves algebraic manipulation to reach a given answer, part (b) requires understanding that P's direction reverses (giving inequality constraints), and part (c) is straightforward energy calculation. While it has multiple parts, each step follows routine procedures taught in FM mechanics with no novel insight required, making it slightly easier than average. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03k Newton's experimental law: direct impact |
3.
A particle $P$ of mass $2 m$ is moving in a straight line with speed $3 u$ on a smooth horizontal table. A second particle $Q$ of mass $3 m$ is moving in the opposite direction to $P$ along the same straight line with speed $u$. The particle $P$ collides directly with $Q$. The direction of motion of $P$ is reversed by the collision. The coefficient of restitution between $P$ and $Q$ is $e$.
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of $Q$ immediately after the collision is $\frac { u } { 5 } ( 8 e + 3 )$
\item Find the range of possible values of $e$.
The total kinetic energy of the particles before the collision is $T$. The total kinetic energy of the particles after the collision is $k T$. Given that $e = \frac { 1 } { 2 }$
\item find the value of $k$.\\[0pt]
[Question 3 Continued]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM Mechanics 2025 Q3 [14]}}